#### Answer

The three real solutions are:
$x = 0,~~$ $~~x=1,~~$ $~~x=6$

#### Work Step by Step

Let's consider a trinomial in this form: $~x^2+bx+c$
To factor this trinomial, we need to find two numbers $r$ and $s$ such that $r+s = b$ and $r\times s = c$
Then the next step is to rewrite the trinomial as follows:
$~x^2+bx+c = (x+r)~(x+s)$
We can use the GCF to factor the given equation:
$4x^3-28x^2+24x = 0$
$(4x)~(x^2-7x+6) = 0$
To factor the left side of this equation, we need to find two numbers $r$ and $s$ such that $r+s = -7~$ and $~r\times s = 6$. We can see that $(-1)+(-6) = -7~$ and $(-1)\times (-6) = 6$
We can solve the equation as follows:
$4x^3-28x^2+24x = 0$
$(4x)~(x^2-7x+6) = 0$
$(4x)~(x-1)~(x-6) = 0$
$4x = 0,~~$ $~~x-1=0,~~$ or $~~x-6=0$
$x = 0,~~$ $~~x=1,~~$ or $~~x=6$